For an equilibrium reaction, `N_(2)O_(4)(g) hArr 2NO_(2)(g)`, the concentrations of `N_(2)O_(4)` and `NO_(2)` at equilibrium are `4.8 xx 10^(-2)` and `1.2 xx 10^(-2) mol//L` respectively. The value of `K_(c)` for the reaction is

Options

A. `3xx 10^(-1) " mol " L^(-1)`
B. `3xx 10^(-3) " mol "L^(-1)`
C. `3xx10^(3) " mol " L^(-1)`
D. `3.3 xx 10^(3) " mol " L^(-1)`

Answer

B

Explanation

`K_(c) =((1.2 xx 10^(-2) "mol "L^(-1))^(2))/((4.8 xx 10^(-2) " mol "L^(-1))) =3.0 xx 10^(-3) "mol "L^(-1)`